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Irrational Exponents! (For rather advanced mathematicians)?
I was thinking about the true nature of exponents last night, and was wondering what exponents really mean. Traditionally, we are told:
A^3 = A x A x A,
A^(3/2) = A^3 x A^(1/2) = A x A x A x root(A)
and so on....
For any rational exponent, this is just fine, as it can be expressed as a fraction, and then done as above.
We give this a meaning... something we can explain, namely:
Integer exponents mean: Multiply this number N amount of times
Roots, or fractional exponents mean: "Which number, multiplied N amount of times, gives you A?"
But for something like A^pi.... What does it MEAN?
We could perhaps say that if B = A^(pi) then ln(B) = (pi)ln(A), and look it up on a log table, or use a Taylor approximation to approximate the ln. For that matter, you could use a Taylor approximation on the original exponent... XD But that doesn't tell us what an irrational exponent really is. I mean, by definition an exponent is the amount of times a number is multiplied by itself. Then we introduced rational exponents, and said that the number at the bottom of the fraction was the root. This still leaves all the irrational numbers out of our understanding of exponents! I know they can be approximated to infinitesimal precision but still. I guess I'm looking for the true definition of A^b.... One that works and makes sense for all real and complex numbers.
Cheers,
M.G.
Woops, you're absolutely right... XD It should be A^(3/2) = (A^3)^1/2 or root(A x A x A)
That's embarrassing.... :s But either way, my question was still fairly clear.
Doing it the way you said:
A^(3) + A^(1/10) + A^(4/100) + A^(1/1000) is an interesting thought... however, that is basically another method for approximating it. I think that if we knew the true meaning, we could go further than just approximate, and perhaps even find numbers that, raised to an irrational number give you a rational number.
For example, e^i(pi) = -1 gives an integer, let alone rational! However, Euler's formula is derived using the taylor expansions of sin, cos and e^x, not with the definition of the exponent. Perhaps with a better definition of exponents, we can get other ones like that.
About the exp(A(lnx))... That one comes from the definition of ln(x), which is an integral at a basic level, that is, one that you have to use Reiman's sum to solve. (ew) and so it's also an approx
2 Answers
- PlogstiesLv 79 years agoFavorite Answer
The expression A^x is defined in analysis to mean exp(A*ln(x)) and this applies to
irrational exponents as well. That is
A^x=exp[A*ln(x)].
- ?Lv 59 years ago
since you're not looking how to calculate, but what it means. You could look at it like this, it's the product of all the decimals
ex:
a^pi=
a^3*a^(1/10)*a^(1/25)*a^(1/1000).... and so on (I'm too lazy to write an infinite amount of decimals.)
remember you add exponents when multiplying.
oh, and one of your exponent rules is wrong.
a^(3/2)=
(a^3)^(1/2)
ex:
(4^3)^(1/2)
64^(1/2)
8
a^3*a^(1/2)
a^(7/2)